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Explore comprehensive NCERT solutions for Class 12 Science Mathematics Chapter 5: Continuity and Differentiability, featuring clear step-by-step explanations. Widely favored by Class 12 Science students, these solutions are invaluable for efficiently completing homework assignments and exam preparation. All questions and answers from Chapter 5 of the NCERT Mathematics textbook for Class 12 Science are presented here at no cost, serving as a valuable resource for students. Page No 159: Question 1: Prove that the functionis continuous at ANSWER: Therefore, f is continuous at x = 0 Therefore, f is continuous at x = −3 Therefore, f is continuous at x = 5 Page No 159: Question 2: Examine the continuity of the function. ANSWER: Thus, f is continuous at x = 3 Page No 159: Question 3: Examine the following functions for continuity. ANSWER: It is evident that f is defined at every real number k and its value at k is k − 5. It is also observed that,  Hence, f is continuous at every real number and therefore, it is a continuous function. (b) The given function is For any real number k ≠ 5, we obtain Hence, f is continuous at every point in the domain of f and therefore, it is a continuous function. (c) The given function is For any real number c ≠ −5, we obtain Hence, f is continuous at every point in the domain of f and therefore, it is a continuous function. (d) The given function is  This function f is defined at all points of the real line. Let c be a point on a real line. Then, c < 5 or c = 5 or c > 5 Case I: c < 5 Then, f (c) = 5 − c Therefore, f is continuous at all real numbers less than 5. Case II : c = 5 Then,  Therefore, f is continuous at x = 5 Case III: c > 5 Therefore, f is continuous at all real numbers greater than 5. Hence, f is continuous at every real number and therefore, it is a continuous function. Page No 159: Question 4: Prove that the function is continuous at x = n, where n is a positive integer. ANSWER: The given function is f (x) = xn It is evident that f is defined at all positive integers, n, and its value at n is nn. Therefore, f is continuous at n, where n is a positive integer. Page No 159: Question 5: Is the function f defined by continuous at x = 0? At x = 1? At x = 2? ANSWER: The given function f is  At x = 0, It is evident that f is defined at 0 and its value at 0 is 0. Therefore, f is continuous at x = 0 At x = 1, f is defined at 1 and its value at 1 is 1. The left hand limit of f at x = 1 is, The right hand limit of f at x = 1 is, Therefore, f is not continuous at x = 1 At x = 2, f is defined at 2 and its value at 2 is 5. Therefore, f is continuous at x = 2 Page No 159: Question 6: Find all points of discontinuity of f, where f is defined by ANSWER: The given function f is It is evident that the given function f is defined at all the points of the real line. Let c be a point on the real line. Then, three cases arise. (i) c < 2 (ii) c > 2 (iii) c = 2 Case (i) c < 2 Therefore, f is continuous at all points x, such that x < 2 Case (ii) c > 2 Therefore, f is continuous at all points x, such that x > 2 Case (iii) c = 2 Then, the left hand limit of f at x = 2 is, The right hand limit of f at x = 2 is, It is observed that the left and right hand limit of f at x = 2 do not coincide. Therefore, f is not continuous at x = 2 Hence, x = 2 is the only point of discontinuity of f. Page No 159: Question 7: Find all points of discontinuity of f, where f is defined by ANSWER: The given function f is The given function f is defined at all the points of the real line. Let c be a point on the real line. Case I: Therefore, f is continuous at all points x, such that x < −3 Case II: Therefore, f is continuous at x = −3 Case III: Therefore, f is continuous in (−3, 3). Case IV: If c = 3, then the left hand limit of f at x = 3 is, The right hand limit of f at x = 3 is, It is observed that the left and right hand limit of f at x = 3 do not coincide. Therefore, f is not continuous at x = 3 Case V: Therefore, f is continuous at all points x, such that x > 3 Hence, x = 3 is the only point of discontinuity of f. Page No 159: Question 8: Find all points of discontinuity of f, where f is defined by ANSWER: The given function f is It is known that, Therefore, the given function can be rewritten as The given function f is defined at all the points of the real line. Let c be a point on the real line. Case I: Therefore, f is continuous at all points x < 0 Case II: If c = 0, then the left hand limit of f at x = 0 is, The right hand limit of f at x = 0 is, It is observed that the left and right hand limit of f at x = 0 do not coincide. Therefore, f is not continuous at x = 0 Case III: Therefore, f is continuous at all points x, such that x > 0 Hence, x = 0 is the only point of discontinuity of f. Page No 159: Question 9: Find all points of discontinuity of f, where f is defined by ANSWER: The given function f is It is known that, Therefore, the given function can be rewritten as Let c be any real number. Then,  Also, Therefore, the given function is a continuous function. Hence, the given function has no point of discontinuity. Page No 159: Question 10: Find all points of discontinuity of f, where f is defined by ANSWER: The given function f is The given function f is defined at all the points of the real line. Let c be a point on the real line. Case I: Therefore, f is continuous at all points x, such that x < 1 Case II: The left hand limit of f at x = 1 is, The right hand limit of f at x = 1 is, Therefore, f is continuous at x = 1 Case III: Therefore, f is continuous at all points x, such that x > 1 Hence, the given function f has no point of discontinuity. Page No 159: Question 11: Find all points of discontinuity of f, where f is defined by ANSWER: The given function f is The given function f is defined at all the points of the real line. Let c be a point on the real line. Case I: Therefore, f is continuous at all points x, such that x < 2 Case II: Therefore, f is continuous at x = 2 Case III: Therefore, f is continuous at all points x, such that x > 2 Thus, the given function f is continuous at every point on the real line. Hence, f has no point of discontinuity. Page No 159: Question 12: Find all points of discontinuity of f, where f is defined by ANSWER: The given function f is The given function f is defined at all the points of the real line. Let c be a point …

NCERT Solutions for Class 12 Science Maths Chapter 5 – Continuity And Differentiability Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Maths Chapter 4 on Determinants, featuring easy-to-follow step-by-step explanations. These solutions have gained immense popularity among Class 12 Science students, serving as a valuable resource for efficiently completing homework assignments and preparing for exams. All the questions and answers from Chapter 4 of the NCERT Book for Class 12 Science Maths are readily available here, providing students with free access to essential study materials. Page No 108: Question 1: Evaluate the determinants in Exercises 1 and 2. ANSWER:  = 2(−1) − 4(−5) = − 2 + 20 = 18 Page No 108: Question 2: Evaluate the determinants in Exercises 1 and 2. (i)  (ii)  ANSWER: (i)  = (cos θ)(cos θ) − (−sin θ)(sin θ) = cos2θ+ sin2θ = 1 (ii)  = (x2 − x + 1)(x + 1) − (x − 1)(x + 1) = x3 − x2 + x + x2 − x + 1 − (x2 − 1) = x3 + 1 − x2 + 1 = x3 − x2 + 2 Page No 108: Question 3: If, then show that ANSWER: The given matrix is. Page No 108: Question 4: If, then show that ANSWER: The given matrix is. It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C1) for easier calculation. From equations (i) and (ii), we have: Hence, the given result is proved. Page No 108: Question 5: Evaluate the determinants (i)  (iii)  (ii)  (iv)  ANSWER: (i) Let. It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation. (ii) Let. By expanding along the first row, we have: (iii) Let By expanding along the first row, we have: (iv) Let By expanding along the first column, we have: Page No 109: Question 6: If, find. ANSWER: Let By expanding along the first row, we have: Page No 109: Question 7: Find values of x, if ANSWER: (i)  (ii)  Page No 109: Question 8: If, then x is equal to (A) 6 (B) ±6 (C) −6 (D) 0 ANSWER: Answer: B Hence, the correct answer is B. Page No 119: Question 1: Using the property of determinants and without expanding, prove that: ANSWER: Page No 119: Question 2: Using the property of determinants and without expanding, prove that: ANSWER: Here, the two rows R1 and R3 are identical. Δ = 0. Page No 119: Question 3: Using the property of determinants and without expanding, prove that: ANSWER: Page No 119: Question 4: Using the property of determinants and without expanding, prove that: ANSWER: By applying C3 → C3 + C2, we have: Here, two columns C1 and C3 are proportional. Δ = 0. Page No 119: Question 5: Using the property of determinants and without expanding, prove that: ANSWER: Applying R2 → R2 − R3, we have: Applying R1 ↔R3 and R2 ↔R3, we have: Applying R1 → R1 − R3, we have: Applying R1 ↔R2 and R2 ↔R3, we have: From (1), (2), and (3), we have: Hence, the given result is proved. Page No 120: Question 6: By using properties of determinants, show that: ANSWER: We have, Here, the two rows R1 and R3 are identical. ∴Δ = 0. Page No 120: Question 7: By using properties of determinants, show that: ANSWER: Applying R2 → R2 + R1 and R3 → R3 + R1, we have: Page No 120: Question 8: By using properties of determinants, show that: (i)  (ii)  ANSWER: (i)  Applying R1 → R1 − R3 and R2 → R2 − R3, we have: Applying R1 → R1 + R2, we have: Expanding along C1, we have: Hence, the given result is proved. (ii) Let. Applying C1 → C1 − C3 and C2 → C2 − C3, we have: Applying C1 → C1 + C2, we have: Expanding along C1, we have: Hence, the given result is proved. Page No 120: Question 9: By using properties of determinants, show that: ANSWER: Applying R2 → R2 − R1 and R3 → R3 − R1, we have: Applying R3 → R3 + R2, we have: Expanding along R3, we have: Hence, the given result is proved. Page No 120: Question 10: By using properties of determinants, show that: (i)  (ii)  ANSWER: (i)  Applying R1 → R1 + R2 + R3, we have: Applying C2 → C2 − C1, C3 → C3 − C1, we have: Expanding along C3, we have: Hence, the given result is proved. (ii)  Applying R1 → R1 + R2 + R3, we have: Applying C2 → C2 − C1 and C3 → C3 − C1, we have: Expanding along C3, we have: Hence, the given result is proved. Page No 120: Question 11: By using properties of determinants, show that: (i)  (ii)  ANSWER: (i)  Applying R1 → R1 + R2 + R3, we have: Applying C2 → C2 − C1, C3 → C3 − C1, we have: Expanding along C3, we have: Hence, the given result is proved. (ii)  Applying C1 → C1 + C2 + C3, we have: Applying R2 → R2 − R1 and R3 → R3 − R1, we have: Expanding along R3, we have: Hence, the given result is proved. Page No 121: Question 12: By using properties of determinants, show that: ANSWER: Applying R1 → R1 + R2 + R3, we have: Applying C2 → C2 − C1 and C3 → C3 − C1, we have: Expanding along R1, we have: Hence, the given result is proved. Page No 121: Question 13: By using properties of determinants, show that: ANSWER: Applying R1 → R1 + bR3 and R2 → R2 − aR3, we have: Expanding along R1, we have: Page No 121: Question 14: By using properties of determinants, show that: ANSWER: Taking out common factors a, b, and c from R1, R2, and R3 respectively, we have: Applying R2 → R2 − R1 and R3 → R3 − R1, we have: Applying C1 → aC1, C2 → bC2, and C3 → cC3, we have: Expanding along R3, we have: Hence, the given result is proved. Page No 121: Question 15: Choose the correct answer. Let A be a square matrix of order 3 × 3, then is equal to A. B. C. D.  ANSWER: Answer: C A is a square matrix of order 3 × 3. Hence, the correct answer is C. Page No 121: Question 16: Which of the following is correct? A. Determinant is a square matrix. B. Determinant is a number associated to a matrix. C. Determinant is a number associated to a square matrix. D. None of these ANSWER: Answer: C We know that to every square matrix, of order n. We can associate a number called the determinant of square matrix A, where element of A. Thus, the determinant is a number associated to a square matrix. Hence, the correct answer is C. Page No …

NCERT Solutions for Class 12 Science Maths Chapter 4 – Determinants Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Mathematics Chapter 3 on Matrices, featuring clear step-by-step explanations. Widely favored by Class 12 Science students, these solutions serve as invaluable resources for completing homework assignments swiftly and preparing for exams effectively. You can access all the questions and answers from Chapter 3 of the NCERT Book for Class 12 Science Mathematics right here, free of charge. Page No 64: Question 1: In the matrix, write: (i) The order of the matrix (ii) The number of elements, (iii) Write the elements a13, a21, a33, a24, a23 ANSWER: (i) In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 × 4. (ii) Since the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements in it. (iii) a13 = 19, a21 = 35, a33 = −5, a24 = 12, a23 =  Page No 64: Question 2: If a matrix has 24 elements, what are the possible order it can have? What, if it has 13 elements? ANSWER: We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24. The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and(6, 4) Hence, the possible orders of a matrix having 24 elements are: 1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4 (1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13. Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1. Page No 64: Question 3: If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements? ANSWER: We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18. The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6,), and (6, 3) Hence, the possible orders of a matrix having 18 elements are: 1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3 (1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5. Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1. Page No 64: Question 4: Construct a 2 × 2 matrix,, whose elements are given by: (i)  (ii)  (iii)  ANSWER: In general, a 2 × 2 matrix is given by (i)  Therefore, the required matrix is (ii)  Therefore, the required matrix is (iii)  Therefore, the required matrix is Page No 64: Question 5: Construct a 3 × 4 matrix, whose elements are given by (i)  (ii)  ANSWER: In general, a 3 × 4 matrix is given by (i)  Therefore, the required matrix is (ii)  Therefore, the required matrix is Page No 64: Question 6: Find the value of x, y, and z from the following equation: (i)  (ii)  (iii)  ANSWER: (i)  As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: x = 1, y = 4, and z = 3 (ii)  As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: x + y = 6, xy = 8, 5 + z = 5 Now, 5 + z = 5 ⇒ z = 0 We know that: (x − y)2 = (x + y)2 − 4xy ⇒ (x − y)2 = 36 − 32 = 4 ⇒ x − y = ±2 Now, when x − y = 2 and x + y = 6, we get x = 4 and y = 2 When x − y = − 2 and x + y = 6, we get x = 2 and y = 4 ∴x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0 (iii)  As the two matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: x + y + z = 9 … (1) x + z = 5 … (2) y + z = 7 … (3) From (1) and (2), we have: y + 5 = 9 ⇒ y = 4 Then, from (3), we have: 4 + z = 7 ⇒ z = 3 ∴ x + z = 5 ⇒ x = 2 ∴ x = 2, y = 4, and z = 3 Page No 64: Question 7: Find the value of a, b, c, and d from the equation: ANSWER: As the two matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: a − b = −1 … (1) 2a − b = 0 … (2) 2a + c = 5 … (3) 3c + d = 13 … (4) From (2), we have: b = 2a Then, from (1), we have: a − 2a = −1 ⇒ a = 1 ⇒ b = 2 Now, from (3), we have: 2 ×1 + c = 5 ⇒ c = 3 From (4) we have: 3 ×3 + d = 13 ⇒ 9 + d = 13 ⇒ d = 4 ∴a = 1, b = 2, c = 3, and d = 4 Page No 65: Question 8: is a square matrix, if (A) m < n (B) m > n (C) m = n (D) None of these ANSWER: The correct answer is C. It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns. Therefore, is a square matrix, if m = n. Page No 65: Question 9: Which of the given values of x and y make the following pair of matrices equal (A)  (B) Not possible to find (C)  (D)  ANSWER: The correct answer is B. It is given that Equating the corresponding elements, we get: We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible. Hence, it is not possible to find the values of x and y for which the given matrices are equal. Page No 65: Question 10: The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is: (A) 27 (B) 18 (C) 81 (D) 512 ANSWER: The correct answer is D. The given matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0 or 1. Now, each of the 9 elements can be filled in two possible ways. Therefore, by the multiplication principle, the required number of possible matrices is 29 = 512 Page No 80: Question 1: Let  Find each of the following (i)  (ii)  (iii)  (iv)  (v)  ANSWER: (i) (ii) (iii) (iv) Matrix A has 2 columns. This number is equal to the number of rows in matrix B. Therefore, AB is defined as: (v) Matrix B has 2 columns. This number is equal to the number of rows in matrix A. Therefore, BA is defined as: Page No 80: Question 2: Compute …

NCERT Solutions for Class 12 Science Maths Chapter 3 – Matrices Read More »

Comprehensive solutions for Class 12 Science Mathematics Chapter 2, Inverse Trigonometric Functions, are available here, complete with straightforward step-by-step explanations. Widely favored among Class 12 Science students, these solutions for Inverse Trigonometric Functions in Mathematics prove invaluable for efficiently completing homework assignments and preparing for exams. All queries and their corresponding answers from Chapter 2 of the NCERT Book for Class 12 Science Mathematics are generously provided on this platform, completely free of charge. Page No 41: Question 1: Find the principal value of  ANSWER: Let sin-1  Then sin y =  We know that the range of the principal value branch of sin−1 is and sin Therefore, the principal value of  Page No 41: Question 2: Find the principal value of  ANSWER: We know that the range of the principal value branch of cos−1 is . Therefore, the principal value of. Page No 41: Question 3: Find the principal value of cosec−1 (2) ANSWER: Let cosec−1 (2) = y. Then,  We know that the range of the principal value branch of cosec−1 is  Therefore, the principal value of  Page No 41: Question 4: Find the principal value of  ANSWER: We know that the range of the principal value branch of tan−1 is  Therefore, the principal value of  Page No 41: Question 5: Find the principal value of  ANSWER: We know that the range of the principal value branch of cos−1 is Therefore, the principal value of  Page No 41: Question 6: Find the principal value of tan−1 (−1) ANSWER: Let tan−1 (−1) = y. Then,  We know that the range of the principal value branch of tan−1 is Therefore, the principal value of  Page No 42: Question 7: Find the principal value of  ANSWER: We know that the range of the principal value branch of sec−1 is Therefore, the principal value of  Page No 42: Question 8: Find the principal value of  ANSWER: We know that the range of the principal value branch of cot−1 is (0,π) and Therefore, the principal value of  Page No 42: Question 9: Find the principal value of  ANSWER: We know that the range of the principal value branch of cos−1 is [0,π] and . Therefore, the principal value of  Page No 42: Question 9: Find the principal value of  ANSWER: We know that the range of the principal value branch of cos−1 is [0,π] and . Therefore, the principal value of  Page No 42: Question 10: Find the principal value of  ANSWER: We know that the range of the principal value branch of cosec−1 is  Therefore, the principal value of  Page No 42: Question 11: Find the value of  ANSWER: Page No 42: Question 12: Find the value of  ANSWER: Page No 42: Question 13: Find the value of if sin−1x = y, then (A) (B)  (C) (D)  ANSWER: It is given that sin−1x = y. We know that the range of the principal value branch of sin−1 is  Therefore,. Page No 42: Question 14: Find the value of is equal to (A) π (B)  (C)  (D)  ANSWER: Page No 47: Question 1: Prove  ANSWER: To prove:  Let x = sinθ. Then,  We have, R.H.S. = = 3θ = L.H.S. Page No 47: Question 2: Prove  ANSWER: To prove: Let x = cosθ. Then, cos−1x =θ. We have, Page No 47: Question 3: Prove  ANSWER: To prove: Page No 47: Question 4: Prove  ANSWER: To prove:  Page No 47: Question 5: Write the function in the simplest form: ANSWER: Page No 47: Question 6: Write the function in the simplest form: ANSWER: Put x = cosec θ ⇒ θ = cosec−1x Page No 47: Question 7: Write the function in the simplest form: ANSWER: Page No 47: Question 8: Write the function in the simplest form: ANSWER: Page No 48: Question 9: Write the function in the simplest form: ANSWER: Page No 48: Question 10: Write the function in the simplest form: ANSWER: Page No 48: Question 11: Find the value of  ANSWER: Let. Then, Page No 48: Question 12: Find the value of  ANSWER: Page No 48: Question 13: Find the value of  ANSWER: Let x = tan θ. Then, θ = tan−1x. Let y = tan Φ. Then, Φ = tan−1y. Page No 48: Question 14: If, then find the value of x. ANSWER: On squaring both sides, we get: Hence, the value of x is Page No 48: Question 15: If, then find the value of x. ANSWER: Page No 48: Question 16: Find the values of  ANSWER: We know that sin−1 (sin x) = x if, which is the principal value branch of sin−1x. Here, Now, can be written as: Page No 48: Question 17: Find the values of  ANSWER: We know that tan−1 (tan x) = x if, which is the principal value branch of tan−1x. Here, Now, can be written as: Page No 48: Question 18: Find the values of  ANSWER: Let. Then, Page No 48: Question 19: Find the values of is equal to (A) (B) (C) (D)  ANSWER: We know that cos−1 (cos x) = x if, which is the principal value branch of cos −1x. Here, Now, can be written as: The correct answer is B. Page No 48: Question 20: Find the values of is equal to (A) (B) (C) (D) 1 ANSWER: Let. Then,  We know that the range of the principal value branch of. ∴ The correct answer is D. Page No 48: Question 21: Find the values of is equal to (A) π (B) (C) 0 (D)  ANSWER: Let. Then, We know that the range of the principal value branch of Let. The range of the principal value branch of The correct answer is B. Page No 51: Question 1: Find the value of  ANSWER: We know that cos−1 (cos x) = x if, which is the principal value branch of cos −1x. Here, Now, can be written as: Page No 51: Question 2: Find the value of  ANSWER: We know that tan−1 (tan x) = x if, which is the principal value branch of tan −1x. Here, Now, can be written as: Page No 51: Question 3: Prove  ANSWER: Now, we have: Page No 51: Question 4: Prove  ANSWER: Now, we have: Page No 51: Question 5: Prove  ANSWER: Now, we will prove that: Page No 51: Question 6: Prove  ANSWER: Now, we have: Page No 51: Question 7: Prove  ANSWER: Using (1) and (2), we have Page No 51: Question 8: Prove  ANSWER: Page No 52: Question 9: Prove  ANSWER: Page No 52: Question 10: Prove  ANSWER: Page No 52: Question 11: Prove  [Hint: putx = cos 2θ] ANSWER: Page No 52: Question 12: Prove  ANSWER: Page No 52: Question 13: Solve ANSWER: Page No 52: Question 14: Solve ANSWER: Page No 52: Question 15: Solveis equal to (A)  (B)  (C)  (D)  …

NCERT Solutions for Class 12 Science Maths Chapter 2 – Inverse Trigonometric Functions Read More »

Discover comprehensive NCERT Solutions for Class 12 Science Mathematics Chapter 1: Relations and Functions. These solutions, complete with easy-to-follow step-by-step explanations, have gained immense popularity among Class 12 Science students. They prove invaluable for swiftly completing homework assignments and effectively preparing for exams. You can access these Maths Relations and Functions Solutions, covering all questions and answers from the NCERT Book of Class 12 Science Mathematics Chapter 1, entirely free of charge. Utilize these resources to enhance your understanding and excel in your academic pursuits. Page No 5: Question 1: Determine whether each of the following relations are reflexive, symmetric and transitive: (i)Relation R in the set A = {1, 2, 3…13, 14} defined as R = {(x, y): 3x − y = 0} (ii) Relation R in the set N of natural numbers defined as R = {(x, y): y = x + 5 and x < 4} (iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x, y): x − y is as integer} (v) Relation R in the set A of human beings in a town at a particular time given by (a) R = {(x, y): x and y work at the same place} (b) R = {(x, y): x and y live in the same locality} (c) R = {(x, y): x is exactly 7 cm taller than y} (d) R = {(x, y): x is wife of y} (e) R = {(x, y): x is father of y} ANSWER: (i) A = {1, 2, 3 … 13, 14} R = {(x, y): 3x − y = 0} ∴R = {(1, 3), (2, 6), (3, 9), (4, 12)} R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R. Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0] Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R. [3(1) − 9 ≠ 0] Hence, R is neither reflexive, nor symmetric, nor transitive. (ii) R = {(x, y): y = x + 5 and x < 4} = {(1, 6), (2, 7), (3, 8)} It is seen that (1, 1) ∉ R. ∴R is not reflexive. (1, 6) ∈R But, (6, 1) ∉ R. ∴R is not symmetric. Now, since there is no pair in R such that (x, y) and (y, z) ∈R, then (x, z) cannot belong to R. ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. (iii) A = {1, 2, 3, 4, 5, 6} R = {(x, y): y is divisible by x} We know that any number (x) is divisible by itself.  (x, x) ∈R ∴R is reflexive. Now, (2, 4) ∈R [as 4 is divisible by 2] But, (4, 2) ∉ R. [as 2 is not divisible by 4] ∴R is not symmetric. Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y. ∴z is divisible by x. ⇒ (x, z) ∈R ∴R is transitive. Hence, R is reflexive and transitive but not symmetric. (iv) R = {(x, y): x − y is an integer} Now, for every x ∈ Z, (x, x) ∈R as x − x = 0 is an integer. ∴R is reflexive. Now, for every x, y ∈ Z if (x, y) ∈ R, then x − y is an integer. ⇒ −(x − y) is also an integer. ⇒ (y − x) is an integer. ∴ (y, x) ∈ R ∴R is symmetric. Now, Let (x, y) and (y, z) ∈R, where x, y, z ∈ Z. ⇒ (x − y) and (y − z) are integers. ⇒ x − z = (x − y) + (y − z) is an integer. ∴ (x, z) ∈R ∴R is transitive. Hence, R is reflexive, symmetric, and transitive. (v) (a) R = {(x, y): x and y work at the same place}  (x, x) ∈ R ∴ R is reflexive. If (x, y) ∈ R, then x and y work at the same place. ⇒ y and x work at the same place. ⇒ (y, x) ∈ R. ∴R is symmetric. Now, let (x, y), (y, z) ∈ R ⇒ x and y work at the same place and y and z work at the same place. ⇒ x and z work at the same place. ⇒ (x, z) ∈R ∴ R is transitive. Hence, R is reflexive, symmetric, and transitive. (b) R = {(x, y): x and y live in the same locality} Clearly (x, x) ∈ R as x and x is the same human being. ∴ R is reflexive. If (x, y) ∈R, then x and y live in the same locality. ⇒ y and x live in the same locality. ⇒ (y, x) ∈ R ∴R is symmetric. Now, let (x, y) ∈ R and (y, z) ∈ R. ⇒ x and y live in the same locality and y and z live in the same locality. ⇒ x and z live in the same locality. ⇒ (x, z) ∈ R ∴ R is transitive. Hence, R is reflexive, symmetric, and transitive. (c) R = {(x, y): x is exactly 7 cm taller than y} Now, (x, x) ∉ R Since human being x cannot be taller than himself. ∴R is not reflexive. Now, let (x, y) ∈R. ⇒ x is exactly 7 cm taller than y. Then, y is not taller than x. ∴ (y, x) ∉R Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x. ∴R is not symmetric. Now, Let (x, y), (y, z) ∈ R. ⇒ x is exactly 7 cm taller thany and y is exactly 7 cm taller than z. ⇒ x is exactly 14 cm taller than z . ∴(x, z) ∉R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. (d) R = {(x, y): x is the wife of y} Now, (x, x) ∉ R Since x cannot be the wife of herself. ∴R is not reflexive. Now, let (x, y) ∈ R ⇒ x is the wife of y. Clearly y is not the wife of x. ∴(y, x) ∉ R Indeed if x is the wife of y, then y is the husband of x. ∴ R is not symmetric. Let (x, y), (y, z) ∈ R ⇒ x is the wife of y and y is the wife of z. This case is not possible. Also, this does not imply that x is the wife of z. ∴(x, z) ∉ R ∴R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. (e) R = {(x, y): x is the father of y} (x, x) ∉ R As x cannot be the father of himself. ∴R is not reflexive. Now, let (x, y) ∈R. ⇒ x is the father of y. ⇒ y cannot be the father of y. Indeed, y is the son or the daughter of y. ∴(y, x) ∉ R ∴ R is not symmetric. Now, let (x, y) ∈ R and (y, z) ∈ R. ⇒ x is the father of y and y is the father of z. ⇒ x is not the father of z. Indeed x is the grandfather of z. ∴ (x, z) ∉ R ∴R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. Page No 5: Question 2: Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive. …

NCERT Solutions for Class 12 Science Maths Chapter 1 – Relations And Functions Read More »

Unlock the Secrets of Class 12 Science Maths Chapter 1: Relations And Functions with our Exclusive NCERT Solutions! Dive into step-by-step explanations tailored for seamless homework completion and exam readiness. Delve into a treasure trove of ad-free resources, only on DD Target PMT NCERT Solutions. Page No 5: Question 1: Determine whether each of the following relations are reflexive, symmetric and transitive: (i)Relation R in the set A = {1, 2, 3…13, 14} defined as R = {(x, y): 3x − y = 0} (ii) Relation R in the set N of natural numbers defined as R = {(x, y): y = x + 5 and x < 4} (iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x, y): x − y is as integer} (v) Relation R in the set A of human beings in a town at a particular time given by (a) R = {(x, y): x and y work at the same place} (b) R = {(x, y): x and y live in the same locality} (c) R = {(x, y): x is exactly 7 cm taller than y} (d) R = {(x, y): x is wife of y} (e) R = {(x, y): x is father of y} ANSWER: (i) A = {1, 2, 3 … 13, 14} R = {(x, y): 3x − y = 0} ∴R = {(1, 3), (2, 6), (3, 9), (4, 12)} R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R. Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0] Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R. [3(1) − 9 ≠ 0] Hence, R is neither reflexive, nor symmetric, nor transitive. (ii) R = {(x, y): y = x + 5 and x < 4} = {(1, 6), (2, 7), (3, 8)} It is seen that (1, 1) ∉ R. ∴R is not reflexive. (1, 6) ∈R But, (6, 1) ∉ R. ∴R is not symmetric. Now, since there is no pair in R such that (x, y) and (y, z) ∈R, then (x, z) cannot belong to R. ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. (iii) A = {1, 2, 3, 4, 5, 6} R = {(x, y): y is divisible by x} We know that any number (x) is divisible by itself.  (x, x) ∈R ∴R is reflexive. Now, (2, 4) ∈R [as 4 is divisible by 2] But, (4, 2) ∉ R. [as 2 is not divisible by 4] ∴R is not symmetric. Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y. ∴z is divisible by x. ⇒ (x, z) ∈R ∴R is transitive. Hence, R is reflexive and transitive but not symmetric. (iv) R = {(x, y): x − y is an integer} Now, for every x ∈ Z, (x, x) ∈R as x − x = 0 is an integer. ∴R is reflexive. Now, for every x, y ∈ Z if (x, y) ∈ R, then x − y is an integer. ⇒ −(x − y) is also an integer. ⇒ (y − x) is an integer. ∴ (y, x) ∈ R ∴R is symmetric. Now, Let (x, y) and (y, z) ∈R, where x, y, z ∈ Z. ⇒ (x − y) and (y − z) are integers. ⇒ x − z = (x − y) + (y − z) is an integer. ∴ (x, z) ∈R ∴R is transitive. Hence, R is reflexive, symmetric, and transitive. (v) (a) R = {(x, y): x and y work at the same place}  (x, x) ∈ R ∴ R is reflexive. If (x, y) ∈ R, then x and y work at the same place. ⇒ y and x work at the same place. ⇒ (y, x) ∈ R. ∴R is symmetric. Now, let (x, y), (y, z) ∈ R ⇒ x and y work at the same place and y and z work at the same place. ⇒ x and z work at the same place. ⇒ (x, z) ∈R ∴ R is transitive. Hence, R is reflexive, symmetric, and transitive. (b) R = {(x, y): x and y live in the same locality} Clearly (x, x) ∈ R as x and x is the same human being. ∴ R is reflexive. If (x, y) ∈R, then x and y live in the same locality. ⇒ y and x live in the same locality. ⇒ (y, x) ∈ R ∴R is symmetric. Now, let (x, y) ∈ R and (y, z) ∈ R. ⇒ x and y live in the same locality and y and z live in the same locality. ⇒ x and z live in the same locality. ⇒ (x, z) ∈ R ∴ R is transitive. Hence, R is reflexive, symmetric, and transitive. (c) R = {(x, y): x is exactly 7 cm taller than y} Now, (x, x) ∉ R Since human being x cannot be taller than himself. ∴R is not reflexive. Now, let (x, y) ∈R. ⇒ x is exactly 7 cm taller than y. Then, y is not taller than x. ∴ (y, x) ∉R Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x. ∴R is not symmetric. Now, Let (x, y), (y, z) ∈ R. ⇒ x is exactly 7 cm taller thany and y is exactly 7 cm taller than z. ⇒ x is exactly 14 cm taller than z . ∴(x, z) ∉R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. (d) R = {(x, y): x is the wife of y} Now, (x, x) ∉ R Since x cannot be the wife of herself. ∴R is not reflexive. Now, let (x, y) ∈ R ⇒ x is the wife of y. Clearly y is not the wife of x. ∴(y, x) ∉ R Indeed if x is the wife of y, then y is the husband of x. ∴ R is not symmetric. Let (x, y), (y, z) ∈ R ⇒ x is the wife of y and y is the wife of z. This case is not possible. Also, this does not imply that x is the wife of z. ∴(x, z) ∉ R ∴R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. (e) R = {(x, y): x is the father of y} (x, x) ∉ R As x cannot be the father of himself. ∴R is not reflexive. Now, let (x, y) ∈R. ⇒ x is the father of y. ⇒ y cannot be the father of y. Indeed, y is the son or the daughter of y. ∴(y, x) ∉ R ∴ R is not symmetric. Now, let (x, y) ∈ R and (y, z) ∈ R. ⇒ x is the father of y and y is the father of z. ⇒ x is not the father of z. Indeed x is the grandfather of z. ∴ (x, z) ∉ R ∴R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. Page No 5: Question 2: Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive. ANSWER: R = {(a, b): a ≤ b2} It can be observed that  ∴R is not reflexive. Now, (1, 4) ∈ R as 1 < 42 But, 4 is not less than 12. ∴(4, 1) ∉ R ∴R is not symmetric. Now, (3, 2), …

NCERT Solutions for Class 12 Science Maths Chapter 1 – Relations And Functions Read More »

Explore the comprehensive NCERT solutions for Class 12 Science Physics Chapter 7 on Communication Systems. These solutions, complete with clear step-by-step explanations, are highly sought after by students. They prove invaluable for completing homework assignments efficiently and preparing for exams. All the questions and answers from the Class 12 Science Physics Chapter 7 in the NCERT Book are available here for free. These Physics Communication Systems Solutions offer a convenient resource for students to grasp concepts and enhance their understanding of the subject. Page No 530: Question 15.1: Which of the following frequencies will be suitable for beyond-the-horizon communication using sky waves? (a) 10 kHz (b) 10 MHz (c) 1 GHz (d) 1000 GHz ANSWER: (b) Answer: 10 MHz For beyond-the-horizon communication, it is necessary for the signal waves to travel a large distance. 10 KHz signals cannot be radiated efficiently because of the antenna size. The high energy signal waves (1GHz − 1000 GHz) penetrate the ionosphere. 10 MHz frequencies get reflected easily from the ionosphere. Hence, signal waves of such frequencies are suitable for beyond-the-horizon communication. Page No 530: Question 15.2: Frequencies in the UHF range normally propagate by means of: (a) Ground waves. (b) Sky waves. (c) Surface waves. (d) Space waves. ANSWER: (d) Answer: Space waves Owing to its high frequency, an ultra high frequency (UHF) wave can neither travel along the trajectory of the ground nor can it get reflected by the ionosphere. The signals having UHF are propagated through line-of-sight communication, which is nothing but space wave propagation. Page No 530: Question 15.3: Digital signals (i) Do not provide a continuous set of values, (ii) Represent values as discrete steps, (iii) Can utilize binary system, and (iv) Can utilize decimal as well as binary systems. Which of the above statements are true? (a) (i) and (ii) only (b) (ii) and (iii) only (c) (i), (ii) and (iii) but not (iv) (d) All of (i), (ii), (iii) and (iv). ANSWER: (c) Answer: A digital signal uses the binary (0 and 1) system for transferring message signals. Such a system cannot utilise the decimal system (which corresponds to analogue signals). Digital signals represent discontinuous values. Page No 530: Question 15.4: Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81m tall. How much service area can it cover if the receiving antenna is at the ground level? ANSWER: Line-of-sight communication means that there is no physical obstruction between the transmitter and the receiver. In such communications it is not necessary for the transmitting and receiving antennas to be at the same height. Height of the given antenna, h = 81 m Radius of earth, R = 6.4 × 106 m For range, d = (2Rh)½, the service area of the antenna is given by the relation: A = πd2 = π (2Rh) = 3.14 × 2 × 6.4 × 106× 81 = 3255.55 × 106 m2 = 3255.55 ∼ 3256 km2 Page No 530: Question 15.5: A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%? ANSWER: Amplitude of the carrier wave, Ac = 12 V Modulation index, m = 75% = 0.75 Amplitude of the modulating wave = Am Using the relation for modulation index: Page No 530: Question 15.6: A modulating signal is a square wave, as shown in Fig. 15.14. The carrier wave is given by  (i) Sketch the amplitude modulated waveform (ii) What is the modulation index? ANSWER: It can be observed from the given modulating signal that the amplitude of the modulating signal, Am = 1 V It is given that the carrier wave c (t) = 2 sin (8πt) ∴Amplitude of the carrier wave, Ac = 2 V Time period of the modulating signal Tm = 1 s The angular frequency of the modulating signal is calculated as: The angular frequency of the carrier signal is calculated as: From equations (i) and (ii), we get: The amplitude modulated waveform of the modulating signal is shown in the following figure. (ii)Modulation index,  Page No 531: Question 15.7: For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2 V. Determine the modulation index μ. What would be the value of μ if the minimum amplitude is zero volt? ANSWER: Maximum amplitude, Amax = 10 V Minimum amplitude, Amin = 2 V Modulation index μ, is given by the relation: Page No 531: Question 15.8: Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station. ANSWER: Let ωc and ωs be the respective frequencies of the carrier and signal waves. Signal received at the receiving station, V = V1 cos (ωc + ωs)t Instantaneous voltage of the carrier wave, Vin = Vc cos ωct At the receiving station, the low-pass filter allows only high frequency signals to pass through it. It obstructs the low frequency signal ωs. Thus, at the receiving station, one can record the modulating signal , which is the signal frequency.

NCERT Solutions for Class 12 Science Physics Chapter 6, titled “Semiconductor Electronics: Materials, Devices And Simple Circuits,” offer comprehensive explanations with step-by-step guidance. These solutions are widely favored by Class 12 Science students for Physics, as they prove invaluable for completing assignments and preparing for exams. All the questions and answers from this chapter in the NCERT Book are provided here for free, aiding students in their understanding of Semiconductor Electronics: Materials, Devices And Simple Circuits. Page No 509: Question 14.1: In an n-type silicon, which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants. ANSWER: The correct statement is (c). In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms. Page No 509: Question 14.2: Which of the statements given in Exercise 14.1 is true for p-type semiconductors. ANSWER: The correct statement is (d). In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms. Page No 509: Question 14.3: Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements is true? (a) (Eg)Si < (Eg)Ge < (Eg)C (b) (Eg)C < (Eg)Ge > (Eg)Si (c) (Eg)C > (Eg)Si > (Eg)Ge (d) (Eg)C = (Eg)Si = (Eg)Ge ANSWER: The correct statement is (c). Of the three given elements, the energy band gap of carbon is the maximum and that of germanium is the least. The energy band gap of these elements are related as: (Eg)C > (Eg)Si > (Eg)Ge Page No 509: Question 14.4: In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them. (b) they move across the junction by the potential difference. (c) hole concentration in p-region is more as compared to n-region. (d) All the above. ANSWER: The correct statement is (c). The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region. Page No 510: Question 14.5: When a forward bias is applied to a p-n junction, it (a) raises the potential barrier. (b) reduces the majority carrier current to zero. (c) lowers the potential barrier. (d) None of the above. ANSWER: The correct statement is (c). When a forward bias is applied to a p-n junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced. Page No 510: Question 14.6: For transistor action, which of the following statements are correct: (a) Base, emitter and collector regions should have similar size and doping concentrations. (b) The base region must be very thin and lightly doped. (c) The emitter junction is forward biased and collector junction is reverse biased. (d) Both the emitter junction as well as the collector junction are forward biased. ANSWER: The correct statement is (b), (c). For a transistor action, the junction must be lightly doped so that the base region is very thin. Also, the emitter junction must be forward-biased and collector junction should be reverse-biased. Page No 510: Question 14.7: For a transistor amplifier, the voltage gain (a) remains constant for all frequencies. (b) is high at high and low frequencies and constant in the middle frequency range. (c) is low at high and low frequencies and constant at mid frequencies. (d) None of the above. ANSWER: The correct statement is (c). The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies. Page No 510: Question 14.8: In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency. ANSWER: Input frequency = 50 Hz For a half-wave rectifier, the output frequency is equal to the input frequency. ∴Output frequency = 50 Hz For a full-wave rectifier, the output frequency is twice the input frequency. ∴Output frequency = 2 × 50 = 100 Hz Page No 510: Question 14.9: For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ. ANSWER: Collector resistance, RC = 2 kΩ = 2000 Ω Audio signal voltage across the collector resistance, V = 2 V Current amplification factor of the transistor, β = 100 Base resistance, RB = 1 kΩ = 1000 Ω Input signal voltage = Vi Base current = IB We have the amplification relation as: Voltage amplification  Therefore, the input signal voltage of the amplifier is 0.01 V. Base resistance is given by the relation: Therefore, the base current of the amplifier is 10 μA. Page No 510: Question 14.13: In an intrinsic semiconductor the energy gap Egis 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration niis given by where n0 is a constant. ANSWER: Energy gap of the given intrinsic semiconductor, Eg = 1.2 eV The temperature dependence of the intrinsic carrier-concentration is written as: Where, kB = Boltzmann constant = 8.62 × 10−5 eV/K T = Temperature n0 = Constant Initial temperature, T1 = 300 K The intrinsic carrier-concentration at this temperature can be written as:  … (1) Final temperature, T2 = 600 K The intrinsic carrier-concentration at this temperature can be written as:  … (2) The ratio between the conductivities at 600 K and at 300 …

NCERT Solutions for Class 12 Science Physics Chapter 6 – Semiconductor Electronics: Materials, Devices And Simple Circuits Read More »

Explore comprehensive NCERT Solutions for Class 12 Science Physics Chapter 5 – Nuclei, featuring clear and concise step-by-step explanations. These solutions have gained immense popularity among Class 12 Science students, proving invaluable for completing homework assignments efficiently and preparing for exams. Free access to all questions and answers from the NCERT Book of Class 12 Science Physics Chapter 5 is available here, ensuring a convenient and effective study resource for students. Page No 462: Question 13.1: (a) Two stable isotopes of lithium  and have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium. (b) Boron has two stable isotopes, and. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of  and . ANSWER: (a) Mass of lithium isotope , m1 = 6.01512 u Mass of lithium isotope , m2 = 7.01600 u Abundance of , η1= 7.5% Abundance of , η2= 92.5% The atomic mass of lithium atom is given as: (b) Mass of boron isotope , m1 = 10.01294 u Mass of boron isotope , m2 = 11.00931 u Abundance of , η1 = x% Abundance of , η2= (100 − x)% Atomic mass of boron, m = 10.811 u The atomic mass of boron atom is given as: And 100 − x = 80.11% Hence, the abundance of  is 19.89% and that of is 80.11%. Page No 462: Question 13.2: The three stable isotopes of neon: and have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon. ANSWER: Atomic mass of , m1= 19.99 u Abundance of , η1 = 90.51% Atomic mass of , m2 = 20.99 u Abundance of , η2 = 0.27% Atomic mass of , m3 = 21.99 u Abundance of , η3 = 9.22% The average atomic mass of neon is given as: Page No 462: Question 13.3: Obtain the binding energy (in MeV) of a nitrogen nucleus, given =14.00307 u ANSWER: Atomic mass of nitrogen, m = 14.00307 u A nucleus of nitrogen  contains 7 protons and 7 neutrons. Hence, the mass defect of this nucleus, Δm = 7mH + 7mn − m Where, Mass of a proton, mH = 1.007825 u Mass of a neutron, mn= 1.008665 u ∴Δm = 7 × 1.007825 + 7 × 1.008665 − 14.00307 = 7.054775 + 7.06055 − 14.00307 = 0.11236 u But 1 u = 931.5 MeV/c2 ∴Δm = 0.11236 × 931.5 MeV/c2 Hence, the binding energy of the nucleus is given as: Eb = Δmc2 Where, c = Speed of light ∴Eb = 0.11236 × 931.5  = 104.66334 MeV Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV. Page No 462: Question 13.4: Obtain the binding energy of the nuclei  and in units of MeV from the following data: = 55.934939 u = 208.980388 u ANSWER: Atomic mass of, m1 = 55.934939 u nucleus has 26 protons and (56 − 26) = 30 neutrons Hence, the mass defect of the nucleus, Δm = 26 × mH + 30 × mn − m1 Where, Mass of a proton, mH = 1.007825 u Mass of a neutron, mn = 1.008665 u ∴Δm = 26 × 1.007825 + 30 × 1.008665 − 55.934939 = 26.20345 + 30.25995 − 55.934939 = 0.528461 u But 1 u = 931.5 MeV/c2 ∴Δm = 0.528461 × 931.5 MeV/c2 The binding energy of this nucleus is given as: Eb1 = Δmc2 Where, c = Speed of light ∴Eb1 = 0.528461 × 931.5  = 492.26 MeV Average binding energy per nucleon  Atomic mass of, m2 = 208.980388 u nucleus has 83 protons and (209 − 83) 126 neutrons. Hence, the mass defect of this nucleus is given as: Δm‘ = 83 × mH + 126 × mn − m2 Where, Mass of a proton, mH = 1.007825 u Mass of a neutron, mn = 1.008665 u ∴Δm‘ = 83 × 1.007825 + 126 × 1.008665 − 208.980388 = 83.649475 + 127.091790 − 208.980388 = 1.760877 u But 1 u = 931.5 MeV/c2 ∴Δm‘ = 1.760877 × 931.5 MeV/c2 Hence, the binding energy of this nucleus is given as: Eb2 = Δm‘c2 = 1.760877 × 931.5 = 1640.26 MeV Average bindingenergy per nucleon =  Page No 462: Question 13.5: A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of atoms (of mass 62.92960 u). ANSWER: Mass of a copper coin, m’ = 3 g Atomic mass of atom, m = 62.92960 u The total number of atoms in the coin Where, NA = Avogadro’s number = 6.023 × 1023atoms /g Mass number = 63 g nucleus has 29 protons and (63 − 29) 34 neutrons ∴Mass defect of this nucleus, Δm‘ = 29 × mH + 34 × mn − m Where, Mass of a proton, mH = 1.007825 u Mass of a neutron, mn = 1.008665 u ∴Δm‘ = 29 × 1.007825 + 34 × 1.008665 − 62.9296 = 0.591935 u Mass defect of all the atoms present in the coin, Δm = 0.591935 × 2.868 × 1022 = 1.69766958 × 1022 u But 1 u = 931.5 MeV/c2 ∴Δm = 1.69766958 × 1022 × 931.5 MeV/c2 Hence, the binding energy of the nuclei of the coin is given as: Eb= Δmc2 = 1.69766958 × 1022 × 931.5  = 1.581 × 1025 MeV But 1 MeV = 1.6 × 10−13 J Eb = 1.581 × 1025 × 1.6 × 10−13 = 2.5296 × 1012 J This much energy is required to separate all the neutrons and protons from the given coin. Page No 462: Question 13.6: Write nuclear reaction equations for (i) α-decay of (ii) α-decay of  (iii) β−-decay of (iv) β−-decay of  (v) β+-decay of (vi) β+-decay of  (vii) Electron capture of  ANSWER: α is a nucleus of helium and β is an electron (e− for β− and e+ for β+). In every α-decay, there is a loss of 2 protons and 4 neutrons. In every β+-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β−-decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus. For the given cases, the various nuclear reactions can be written as: Page No 462: Question 13.7: A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value? ANSWER: Half-life of the radioactive isotope = T years Original amount of the radioactive isotope = N0 (a) After decay, the amount of the radioactive isotope = N It is given that only 3.125% of N0 remains after decay. Hence, we can write: Where, λ = Decay constant t = Time Hence, the isotope will take about 5T years to reduce to 3.125% of its original value. (b) After decay, the amount of the radioactive isotope = N It …

NCERT Solutions for Class 12 Science Physics Chapter 5 – Nuclei Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Physics Chapter 4 on Atoms, featuring clear, step-by-step explanations. Widely favored by class 12 Science students, these Physics Atoms Solutions serve as a valuable resource for efficiently completing homework assignments and preparing for examinations. You can access all the questions and answers from the NCERT Book of Class 12 Science Physics Chapter 4 at no cost. These solutions are designed to assist you in grasping the concepts quickly and effectively, making your study sessions more productive. Page No 435: Question 12.1: Choose the correct alternative from the clues given at the end of the each statement: (a) The size of the atom in Thomson’s model is ………. the atomic size in Rutherford’s model. (much greater than/no different from/much less than.) (b) In the ground state of ………. electrons are in stable equilibrium, while in ………. electrons always experience a net force. (Thomson’s model/ Rutherford’s model.) (c) A classical atom based on ………. is doomed to collapse. (Thomson’s model/ Rutherford’s model.) (d) An atom has a nearly continuous mass distribution in a ………. but has a highly non-uniform mass distribution in ………. (Thomson’s model/ Rutherford’s model.) (e) The positively charged part of the atom possesses most of the mass in ………. (Rutherford’s model/both the models.) ANSWER: (a) The sizes of the atoms taken in Thomson’s model and Rutherford’s model have the same order of magnitude. (b) In the ground state of Thomson’s model, the electrons are in stable equilibrium. However, in Rutherford’s model, the electrons always experience a net force. (c) A classical atom based on Rutherford’s model is doomed to collapse. (d) An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model. (e) The positively charged part of the atom possesses most of the mass in both the models. Page No 435: Question 12.2: Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect? ANSWER: In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 × 10−27 kg) is less than the mass of incident α−particles (6.64 × 10−27 kg). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α−particles would not bounce back if solid hydrogen is used in the α-particle scattering experiment. Page No 436: Question 12.3: What is the shortest wavelength present in the Paschen series of spectral lines? ANSWER: Rydberg’s formula is given as: Where, h = Planck’s constant = 6.6 × 10−34 Js c = Speed of light = 3 × 108 m/s (n1 and n2 are integers) The shortest wavelength present in the Paschen series of the spectral lines is given for values n1 = 3 and n2 = ∞. Page No 436: Question 12.4: A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level? ANSWER: Separation of two energy levels in an atom, E = 2.3 eV = 2.3 × 1.6 × 10−19 = 3.68 × 10−19 J Let ν be the frequency of radiation emitted when the atom transits from the upper level to the lower level. We have the relation for energy as: E = hv Where, h = Planck’s constant Hence, the frequency of the radiation is 5.6 × 1014 Hz. Page No 436: Question 12.5: The ground state energy of hydrogen atom is −13.6 eV. What are the kinetic and potential energies of the electron in this state? ANSWER: Ground state energy of hydrogen atom, E = − 13.6 eV This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy. Kinetic energy = − E = − (− 13.6) = 13.6 eV Potential energy is equal to the negative of two times of kinetic energy. Potential energy = − 2 × (13.6) = − 27 .2 eV Page No 436: Question 12.6: A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of the photon. ANSWER: For ground level, n1 = 1 Let E1 be the energy of this level. It is known that E1 is related with n1 as: The atom is excited to a higher level, n2 = 4. Let E2 be the energy of this level. The amount of energy absorbed by the photon is given as: E = E2 − E1 For a photon of wavelengthλ, the expression of energy is written as: Where, h = Planck’s constant = 6.6 × 10−34 Js c = Speed of light = 3 × 108 m/s And, frequency of a photon is given by the relation, Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 × 1015 Hz. Page No 436: Question 12.7: (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels. ANSWER: (a) Let ν1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1. For charge (e) of an electron, ν1 is given by the relation, Where, e = 1.6 × 10−19 C ∈0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2 h = Planck’s constant = 6.62 × 10−34 Js For level n2 = 2, we can write the relation for the corresponding orbital speed as: And, for n3 = 3, we can write the relation for the corresponding orbital speed as: Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 × 106 m/s, 1.09 × 106 m/s, 7.27 × 105 m/s respectively. (b) Let T1 be the orbital period of the electron when it is in level n1 = 1. Orbital period is related to orbital speed as: Where, r1 = Radius of the orbit h = Planck’s constant = 6.62 × 10−34 Js e = Charge on an electron = 1.6 × 10−19 C ∈0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2 m = Mass …

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